KCET · Physics · Nuclear Physics
On bombarding \(\mathrm{U}^{235}\) by slow neutron, \(200 \mathrm{MeV}\) energy is released. If the power output of atomic reactor is \(1.6 \mathrm{MW}\), then the rate of fission will be
- A \(5 \times 10^{22} / \mathrm{s}\)
- B \(5 \times 10^{16} / \mathrm{s}\)
- C \(8 \times 10^{16} / \mathrm{s}\)
- D \(20 \times 10^{16} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(5 \times 10^{16} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Energy released on bombarding \(\mathrm{U}^{235}\) by neutron \(=200 \mathrm{MeV}\)
Power output of atomic reactor \(=1.6 \mathrm{MW}\)
\(\therefore\) Rate of fission \(=\frac{1.6 \times 10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\)
\[
=5 \times 10^{16} / \mathrm{s}
\]
Power output of atomic reactor \(=1.6 \mathrm{MW}\)
\(\therefore\) Rate of fission \(=\frac{1.6 \times 10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\)
\[
=5 \times 10^{16} / \mathrm{s}
\]
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